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sertanlavr [38]
3 years ago
12

For the oscillating bob of a pendulum, kinetic energy is at its minimum at the maximum height. Which statement is true, accordin

g to the Work-Energy theorem?
A) Maximum work is done on the bob between points A and C.
B) Minimum work is done on the bob between points B and C.
C) Maximum work is done on the bob between points A and B.
D) Minimum work is done on the bob between points A and B.

Physics
1 answer:
BabaBlast [244]3 years ago
5 0

Answer:

C) Maximum work is done on the bob between points A and B.

Explanation:

The work done converts all potential energy to kinetic energy

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Distinguish between endangered and extinct species.
Julli [10]

Answer and Explanation:

The distinction between Endangered species and Extict species is Given as under:

Endangered Species:

Endangered species are categorized as those species of flora and fauna which are on the verge of extinction if not provided timely protection in order to save them from getting extinct.

Some of the endangered species are African Penguin, Bengal Tiger, Blue Whale, etc

Extinct Species:

Those species of Flora and Fauna which have vanished completely or either from the wild i.e., that individuals of that species are not alive any more and have completely disappear from the earth.

In practice, if a species is not seen for over a period of 50 years it is considered to be extinct.

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7 0
3 years ago
For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is
ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)

where A(t)=A₀e^{\frac{-bt}{2m}}

 A₀ is the amplitude at t=0 and

w' is the angular frequency of damped SHM, which is given by,

w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

Now coming to the problem,

Given: m=1.2 kg

           k=9.8 N/m

           b=210 g/s= 0.21 kg/s

           A₀=13 cm

a) A(t)=A₀/8

⇒A₀e^{\frac{-bt}{2m}} =A₀/8

⇒e^{\frac{bt}{2m}}=8

applying logarithm on both sides

⇒\frac{bt}{2m}=ln(8)

⇒t=\frac{2m*ln(8)}{b}

substituting the values

t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)

b) w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}

T'=\frac{2\pi}{w'}, where T' is time period of damped SHM

⇒T'=\frac{2\pi}{2.86}=2.2s

let n be number of oscillations made

then, nT'=t

⇒n=\frac{24}{2.2}=11(approx)

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4 years ago
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WINSTONCH [101]
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a car traveling in a highway brakes to a stop without skidding. what happens to its kinetic energy? can that energy be recovered
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Answer:

A car traveling on a highway takes to stop without skidding. What happens to its kinetic energy? ... The energy goes to parts of the brakes. No, the energy can only be reused if there is a heat pump present

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bagirrra123 [75]

Answer:

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Explanation:

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